Q#9 (Past Exam Paper – June 2016 Paper 22 Q1)
(a) Define acceleration. [1]
(b) A man travels on a toboggan down a slope covered with snow from point A to point B and then to point C. The path is illustrated in Fig. 1.1.
Fig. 1.1 (not to scale)
The slope AB makes an angle of 40° with the horizontal and the slope BC makes an angle of 20° with the horizontal. Friction is not negligible.
The man and toboggan have a combined mass of 95 kg.
The man starts from rest at A and has constant acceleration between A and B. The man
takes 19 s to reach B. His speed is 36 m s-1 at B.
(i) Calculate the acceleration from A to B. [2]
(ii) Show that the distance moved from A to B is 340 m. [1]
(iii) For the man and toboggan moving from A to B, calculate
1. the change in kinetic energy, [2]
2. the change in potential energy. [2]
(iv) Use your answers in (iii) to determine the average frictional force that acts on the
toboggan between A and B. [2]
(v) A parachute opens on the toboggan as it passes point B. There is a constant deceleration of 3.0 m s-2 from B to C.
Calculate the frictional force that produces this deceleration between B and C. [2]
[Total: 12]
Solution:
(a)
Acceleration is defined as the rate of change of velocity.
(b)
(i)
{v = u + at where u = 0 (at rest)}
v = 0 + at or v = at
a = 36 / 19 = 1.9 (1.8947) m s-2
(ii)
{s = average speed × time v2 = u2 + 2as s = ut + ½ at2 }
s = ½(u + v)t or s = v2 / 2a or s = ½ at2
= ½×36×19 = 362 / (2×1.89) = ½×1.89×192
= 340 m (342 m / 343 m / 341 m)
(iii)
1.
{ΔKE = ½ m (v2 – u2)}
ΔKE = ½×95×(362 – 02)
= 62 000 (61 560) J
2.
{ΔPE = mgΔh where Δh is the vertical height = 340 sin 40°}
ΔPE = 95×9.81×340 sin 40° or 95×9.81×218.5
= 200 000 J
(iv)
{As the toboggan moves down the slope,
PE at A = KE at B + Work done against friction}
work done (by frictional force) = ΔPE – ΔKE
or
work done = 200 000 – 62 000 (values from 1b(iii) 1. and 2.)
{Work done = 138 000 N
Work done = (Frictional) Force × distance}
Frictional force = (138 000 / 340 =) 410 (406) N
(v)
{Since there is a deceleration, the resultant force is upwards.
Resultant force = Frictional force – Component of Weight along slope
OR – Resultant force = Component of Weight along slope – Frictional force
Component of Weight along slope = mg sin 20°}
–ma = mg sin 20° – f
or ma = –mg sin 20° + f
–95 × 3.0 = 95 × 3.36 – f
Frictional force f = 600 (604) N
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